Page 589 - Advanced_Engineering_Mathematics o'neil

16.2 Wave Motion on an Interval 569

So far, for (n = 1,2,···) we have a function

y n (x,t) = c n sin(nπx/L)cos(nπct/L)

that satisfies the wave equation, both boundary conditions and the initial condition of zero initial
velocity. We have yet to satisfy y(x,0) = f (x).If f (x) = K sin(mπx/L) for some integer m,
then y(x,t)= K sin(mπx/L)cos(mπct/L) is the solution. But f (x) need not look like this. For
example, if the string is picked up at its midpoint, say

x for 0 ≤ x ≤ L/2
f (x) = (16.4)
L − x for L/2 ≤ x ≤ L
N
then not even a finite sum y(x,t) = y n (x,t) can satisfy y(x,0) = f (x). In such a case we
n=0
use an infinite superposition
∞

nπx nπct
y(x,t) = c n sin cos .
L L
n=1
The condition y(x,0) = f (x) is satisfied if we can choose the coefficients so that
∞

nπx
y(x,0) = f (x) = c n sin .
L
n=0
This is the Fourier sine expansion of f (x) on [0, L], hence choose
2 L nπξ
c n = f (ξ)sin dξ.
L 0 L
The solution of the problem, with zero initial velocity and initial position given by f ,is

2
L nπx nπct
∞
y(x,t) = f (ξ)sin(nπξ/L)dξ sin cos . (16.5)
L 0 L L
n=1
EXAMPLE 16.1
With f given by equation (16.4) and L = π, the coefficients are
2 π
c n = f (ξ)sin(nξ)dξ
π 0
2 π/2 2 π
= ξ sin(nξ)dξ + (π − ξ)sin(nξ)dξ
π 0 π π/2
4sin(nπ/2)
= .
πn 2
The solution of this problem is
∞

4sin(nπ/2)
y(x,t) = sin(nx)cos(nct).
πn 2
n=1
Figure 16.2 shows the string profile for c = 2 at times t = 0.3,0.6, 0.9, and 1.2, starting at
the top and moving downward in this time frame.

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October 14, 2010 15:23 THM/NEIL Page-569 27410_16_ch16_p563-610